Problem 52 - Precalculus: Mathematics for Calculus

Question

Show that the given statement is true. $$\left(\begin{array}{l}n \\ r\end{array}\right)=\left(\begin{array}{c}n \\\ n-r\end{array}\right) \quad\( for \)0 \leq r \leq n$$

Step-by-Step Solution

Answer

The given statement is true because binomial coefficients have a symmetry property: \( \binom{n}{r} = \binom{n}{n-r} \).

Step 1: Understand the Problem Statement

The problem asks us to prove that the binomial coefficient \( \binom{n}{r} \) is equal to \( \binom{n}{n-r} \). This is a known property of binomial coefficients and is based on the symmetric property of combinations.

Step 2: Define the Binomial Coefficient

The binomial coefficient \( \binom{n}{r} \) is defined as the number of ways to choose \( r \) elements from \( n \) elements without regard to order. It is calculated using the formula: \[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \] where \( n! \) denotes the factorial of \( n \).

Step 3: Express Both Sides of the Equation

Using the definition of a binomial coefficient, express the right side of the equation: \( \binom{n}{n-r} \). By the definition, this becomes: \[ \binom{n}{n-r} = \frac{n!}{(n-r)!r!} \]. Notice that this expression is the same as \( \binom{n}{r} \).

Step 4: Verify the Symmetry Property

Since \( \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!r!} \), the expressions are identical. This is because multiplication is commutative, so \( r!(n-r)! = (n-r)!r! \), confirming the binomial coefficient property.

Step 5: Conclusion

We have verified that \( \binom{n}{r} = \binom{n}{n-r} \) by showing both expressions are identical. This proves the symmetry property of binomial coefficients as required.

Key Concepts

Binomial Coefficient

The binomial coefficient is a fundamental concept in combinatorics that deals with the number of ways to choose a subset of items from a larger set. When we talk about the binomial coefficient, we write it as \( \binom{n}{r} \), which is read as '\( n \) choose \( r \)'.
It represents the number of ways to select \( r \) elements from a total of \( n \) elements without considering the order of selection. The formula to calculate this is:

• \( \binom{n}{r} = \frac{n!}{r!(n-r)!} \)

Here, \( n! \) is a factorial, which means you multiply all whole numbers from \( n \) down to 1.
For example, if you have 5 people and want to choose 2 to form a committee, you would compute \( \binom{5}{2} \). This simple calculation lets you determine all possible combinations for choosing items from a group without needing to list them all.

Symmetry Property

The symmetry property of binomial coefficients is a fascinating and important feature in combinatorics. It states that

• \( \binom{n}{r} = \binom{n}{n-r} \)

This property demonstrates that choosing \( r \) elements from \( n \) is equivalent to choosing \( n-r \) elements to exclude from \( n \).
The logic here is simple and intuitive: selecting a subset of \( r \) is the same as leaving out \( n-r \). This means there are exactly the same number of ways to do either task.
This property is very useful, particularly for simplifying complex combinatorial expressions, and to verify calculations when symmetry is recognized. In the context of a binomial expansion, it's a key reason why binomial coefficients are mirror-symmetric.

Factorial

Factorial is a mathematical operation that plays a crucial role in calculating binomial coefficients. A factorial, expressed as \( n! \), is the product of all positive integers less than or equal to \( n \). It’s important for understanding permutations and combinations.
Here’s how you calculate a factorial:

• \( n! = n \times (n-1) \times (n-2) \times \cdots \times 1 \)

Factorials grow very rapidly, which makes them a powerful tool in counting and arrangement problems. For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
In the context of the binomial coefficient, factorials are used in the denominator and numerator of the formula to account for total possible arrangements and to divide by the number of redundant permutations.
This ensures that when determining a combination, order does not matter, allowing mathematicians and students to focus on counting distinct selections.